Problem: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{-4a^3 + 4a}{9a^3 + 36a^2 + 27a}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {-4a(a^2 - 1)} {9a(a^2 + 4a + 3)} $ $ r = -\dfrac{4a}{9a} \cdot \dfrac{a^2 - 1}{a^2 + 4a + 3} $ Simplify: $ r = - \dfrac{4}{9} \cdot \dfrac{a^2 - 1}{a^2 + 4a + 3}$ Since we are dividing by $a$ , we must remember that $a \neq 0$ Next factor the numerator and denominator. $ r = - \dfrac{4}{9} \cdot \dfrac{(a + 1)(a - 1)}{(a + 1)(a + 3)}$ Assuming $a \neq -1$ , we can cancel the $a + 1$ $ r = - \dfrac{4}{9} \cdot \dfrac{a - 1}{a + 3}$ Therefore: $ r = \dfrac{ -4(a - 1)}{ 9(a + 3)}$, $a \neq -1$, $a \neq 0$